parsimonious squares
 

11^2= 121
111^2= 12321
36361^2= 1322122321
363639^2= 132233322321
461761^2= 213223221121
3636361^2= 13223121322321
34815389^2= 1212111311221321
362397739^2= 131332121232312121

finitely many such numbers? infinitely many if you allow four kinds of digit?

eg 99983333^2=9996666877788889

if you allow {7,8,9} as the digits, looks as if only 3^2=9
{6,7,8} looks like only 26^2=676

Hey, need a precise definition...otherwise:
11^2 = 121
101^2 = 10201
1001^2 = 1002001
...... and I've only got three digits anywhere in my relations, and we can change the base to anything we like 3 or greater....

you're using digits 0, 1, 2.

i was using digits 1, 2, 3.

obviously (10^n)^2 uses only 0,1 but that's boring

[QUOTE=fivemack;264776]you're using digits 0, 1, 2.

i was using digits 1, 2, 3.

obviously (10^n)^2 uses only 0,1 but that's boring[/QUOTE]

[url]http://www.google.ca/search?hl=en&q=parsimonious&tbs=dfn:1&tbo=u&sa=X&ei=KsgITvbLFYecgQffzNnpAg&ved=0CBcQkQ4&biw=1066&bih=749[/url] either stingy or as easy as possible is what I find.

I suppose you've checked A30174 and A30175...
I think if the sequence is finite, then also a sequence with four digits is finite unless there exists an infinite pattern like the one Christenson pointed out.

[QUOTE=science_man_88;264777][url]http://www.google.ca/search?hl=en&q=parsimonious&tbs=dfn:1&tbo=u&sa=X&ei=KsgITvbLFYecgQffzNnpAg&ved=0CBcQkQ4&biw=1066&bih=749[/url] either stingy or as easy as possible is what I find.[/QUOTE]

"Stingy", "frugal" is the meaning here. The squares don't "use" very many different digits.

[QUOTE=Mr. P-1;264791]"Stingy", "frugal" is the meaning here. The squares don't "use" very many different digits.[/QUOTE]

Let's post working definitions:

A square(or cube) is said to be parsimonious if it contains many fewer digits than the number base it is expressed in.

That is, it is parsimonious of order 2 if it only contains 2 distinct digits, such as 49.

Note that given a sufficiently large number base to express things in, any square is parismonious since it can be represented as 100 in the base of the square root.

A square is ultraparsimonious if both the square and its root are parsimonious in the same way.

Fivemack's original question: Are there infinitely many parsimonious squares, base 10, using only digits {1,2,3}?

Spuriously, I conjecture that a procedure exists, given an arbitrary sequence of digits from {1,2,3} to construct a perfect square by appending additional digits {1,2,3}. It has something to do with the 832 year cycle with too many Fridays in July. :devil:

It vaguely works (?) like this:
Assume 2 digits will be added. Choose among the 9 allowed combinations of digits to minimize the distance to a perfect square.
Repeat (832 times, remember we need an arbitrarily long sequence) until the distance is zero.

'Distance to a perfect square' is quite difficult to define here

I think you probably want to work p-adically backwards from the end

but you're having to branch quite a lot, and it's always possible that all the branches will fall off

(for example about 10^-25 of numbers of 50 digits are squares, and 3^50 numbers of 50 digits use only the digits 123, and 3^50/10^25 is significantly <1)

[QUOTE=Christenson;264729]Hey, need a precise definition...otherwise:
11^2 = 121
101^2 = 10201
1001^2 = 1002001
...... and I've only got three digits anywhere in my relations, and we can change the base to anything we like 3 or greater....[/QUOTE]

[QUOTE=fivemack;264776]you're using digits 0, 1, 2.

i was using digits 1, 2, 3.

obviously (10^n)^2 uses only 0,1 but that's boring[/QUOTE]

[QUOTE=mart_r;264787]I suppose you've checked A30174 and A30175...
I think if the sequence is finite, then also a sequence with four digits is finite unless there exists an infinite pattern like the one Christenson pointed out.[/QUOTE]

To make such an effort more interesting, I would suggest not allowing the digit of 0.

[QUOTE=fivemack;264721]11^2= 121
111^2= 12321
36361^2= 1322122321
363639^2= 132233322321
461761^2= 213223221121
3636361^2= 13223121322321
34815389^2= 1212111311221321
362397739^2= 131332121232312121

finitely many such numbers? infinitely many if you allow four kinds of digit?

eg 99983333^2=9996666877788889

if you allow {7,8,9} as the digits, looks as if only 3^2=9
{6,7,8} looks like only 26^2=676[/QUOTE]

Based on what is being allowed 1^2= 1 would also fit the {1,2,3} pattern.

[QUOTE=Christenson;264795]Let's post working definitions:

A square(or cube) is said to be parsimonious if it contains many fewer digits than the number base it is expressed in.

That is, it is parsimonious of order 2 if it only contains 2 distinct digits, such as 49.

Note that given a sufficiently large number base to express things in, any square is parismonious since it can be represented as 100 in the base of the square root.

A square is ultraparsimonious if both the square and its root are parsimonious in the same way.

Fivemack's original question: Are there infinitely many parsimonious squares, base 10, using only digits {1,2,3}?

Spuriously, I conjecture that a procedure exists, given an arbitrary sequence of digits from {1,2,3} to construct a perfect square by appending additional digits {1,2,3}. It has something to do with the 832 year cycle with too many Fridays in July. :devil:

It vaguely works (?) like this:
Assume 2 digits will be added. Choose among the 9 allowed combinations of digits to minimize the distance to a perfect square.
Repeat (832 times, remember we need an arbitrarily long sequence) until the distance is zero.[/QUOTE]


It's an annoying question for me to try, but I know that given base ABC ( digits)C must be 1 or 9 but it all depends on the carry what B can be.